Given The Nuclear Equation $ {}_{94}^{239} Pu \longrightarrow X+{ }_2^4 He $, What Is The Element X?
In the realm of nuclear chemistry, understanding nuclear equations is paramount to deciphering the transformations of atomic nuclei. These equations, symbolic representations of nuclear reactions, adhere to fundamental conservation laws, ensuring the balance of mass and atomic numbers. This article delves into the analysis of a specific nuclear equation, focusing on identifying the unknown element "X" produced during the decay of Plutonium-239.
Dissecting the Nuclear Equation
The nuclear equation presented is:
$ {}_{94}^{239} Pu \longrightarrow X + {}_2^4 He $
This equation depicts the alpha decay of Plutonium-239 ($ {}_{94}^{239} Pu {}_2^4 He $), which consists of two protons and two neutrons, essentially a helium nucleus. The process results in the transformation of the parent nucleus (Plutonium-239 in this case) into a daughter nucleus (represented by "X") and the emission of the alpha particle.
To identify the element X, we must apply the fundamental principles of nuclear equations, which revolve around the conservation of mass number (the superscript) and atomic number (the subscript). The mass number represents the total number of protons and neutrons in the nucleus, while the atomic number signifies the number of protons, defining the element's identity.
Conservation Laws in Action
The conservation of mass number dictates that the sum of the mass numbers on the left side of the equation must equal the sum of the mass numbers on the right side. Similarly, the conservation of atomic number requires that the sum of the atomic numbers on both sides of the equation remains constant. Applying these principles to our equation:
- Mass Number Conservation: 239 (Pu) = A (X) + 4 (He)
- Atomic Number Conservation: 94 (Pu) = Z (X) + 2 (He)
Where A represents the mass number of element X, and Z represents its atomic number. Solving these simple equations will reveal the values of A and Z, which will then allow us to identify the element X.
Solving for A and Z
From the mass number conservation equation, we can isolate A:
A = 239 - 4 = 235
Similarly, from the atomic number conservation equation, we can solve for Z:
Z = 94 - 2 = 92
Therefore, element X has a mass number of 235 and an atomic number of 92. Now, we need to consult the periodic table to identify the element with an atomic number of 92.
Identifying Element X
The periodic table is the quintessential resource for identifying elements based on their atomic numbers. It organizes elements in ascending order of their atomic numbers, providing a clear mapping between the atomic number and the element's symbol.
By consulting the periodic table, we find that the element with an atomic number of 92 is Uranium (U). Therefore, element X is Uranium.
Specifying the Isotope
We've determined that element X is Uranium, but we also need to specify the isotope. Isotopes are variants of an element that have the same atomic number but different mass numbers due to varying numbers of neutrons. In this case, element X has a mass number of 235, so it is Uranium-235 ($ {}_{92}^{235} U $).
Therefore, the complete nuclear equation is:
$ {}{94}^{239} Pu \longrightarrow {}{92}^{235} U + {}_2^4 He $
This equation illustrates the alpha decay of Plutonium-239 into Uranium-235, with the emission of an alpha particle.
Conclusion
By meticulously applying the principles of conservation of mass number and atomic number, we successfully identified element X as Uranium-235 ($ {}_{92}^{235} U $). This exercise underscores the importance of understanding nuclear equations and the fundamental laws that govern nuclear transformations. The ability to decipher these equations is crucial in various fields, including nuclear chemistry, nuclear physics, and nuclear engineering, enabling scientists and engineers to comprehend and manipulate nuclear processes.
Alpha decay is a fascinating nuclear process where an unstable atomic nucleus ejects an alpha particle. This transformation is driven by the inherent instability of certain heavy nuclei, which contain an excess of protons and neutrons. The emission of an alpha particle, a tightly bound cluster of two protons and two neutrons, allows the nucleus to move towards a more stable configuration. To grasp the significance of alpha decay, we must delve into the underlying reasons for nuclear instability and the characteristics of the alpha particle itself.
Nuclear Instability and the Strong Nuclear Force
The atomic nucleus is a realm of intense forces. Protons, with their positive charges, exert repulsive electrostatic forces on each other. These repulsive forces, if unchecked, would cause the nucleus to fly apart. However, the nucleus remains bound together due to the strong nuclear force, a fundamental force of nature that acts between nucleons (protons and neutrons). The strong nuclear force is attractive and far more potent than the electrostatic force, but it operates over very short distances.
The balance between the electrostatic repulsion and the strong nuclear force determines the stability of a nucleus. In lighter nuclei, the number of neutrons and protons is roughly equal, providing sufficient strong nuclear force to overcome the electrostatic repulsion. However, as the nucleus becomes heavier, the number of protons increases, leading to a greater electrostatic repulsion. To compensate for this increased repulsion, heavier nuclei require a greater proportion of neutrons. Neutrons contribute to the strong nuclear force without adding to the electrostatic repulsion, effectively diluting the concentration of protons.
However, there is a limit to how many neutrons can be added to stabilize a nucleus. Beyond a certain point, the nucleus becomes too large and the strong nuclear force, with its short-range nature, can no longer effectively counteract the electrostatic repulsion. This leads to nuclear instability and the propensity for radioactive decay, including alpha decay.
The Alpha Particle A Helium Nucleus
The alpha particle ($ {}_2^4 He $) is a particularly stable configuration of nucleons. It consists of two protons and two neutrons, forming a helium nucleus. This specific combination of nucleons results in a highly bound structure with a large binding energy. The binding energy is the energy required to disassemble the nucleus into its constituent protons and neutrons. A high binding energy signifies a stable nucleus.
The alpha particle's stability is a key factor in alpha decay. When a heavy nucleus emits an alpha particle, it effectively sheds four nucleons in a highly stable configuration. This significantly reduces the mass and atomic number of the nucleus, moving it closer to the belt of stability on the chart of nuclides, which plots the stable isotopes of elements.
The Process of Alpha Decay
The process of alpha decay can be visualized as follows: Within the unstable nucleus, an alpha particle momentarily forms due to random fluctuations in the positions and energies of the nucleons. This alpha particle is trapped within the nucleus by the strong nuclear force. However, the alpha particle also experiences the repulsive electrostatic force from the protons in the nucleus. This creates a potential barrier that the alpha particle must overcome to escape the nucleus.
According to classical physics, the alpha particle should not be able to overcome this potential barrier unless it possesses sufficient energy. However, quantum mechanics allows for a phenomenon called quantum tunneling, where a particle can pass through a potential barrier even if it does not have enough energy to surmount it classically. The probability of tunneling depends on the height and width of the potential barrier and the energy of the alpha particle.
In alpha decay, the alpha particle tunnels through the potential barrier and escapes the nucleus. The emitted alpha particle carries away kinetic energy, which is manifested as its high speed. The daughter nucleus, now with two fewer protons and two fewer neutrons, recoils in the opposite direction to conserve momentum.
Consequences of Alpha Decay
Alpha decay has several significant consequences:
- Change in Atomic Number: The emission of an alpha particle reduces the atomic number of the nucleus by two. This means that the element transforms into a different element. For instance, Plutonium (atomic number 94) transforms into Uranium (atomic number 92) via alpha decay.
- Change in Mass Number: Alpha decay also reduces the mass number of the nucleus by four, as the alpha particle carries away four nucleons.
- Energy Release: Alpha decay is an exothermic process, meaning it releases energy. This energy is primarily manifested as the kinetic energy of the alpha particle and the recoiling daughter nucleus.
- Radioactivity: Alpha decay is a form of radioactivity, and alpha particles are a type of ionizing radiation. This means that alpha particles can interact with matter and knock electrons out of atoms, creating ions. Ionizing radiation can be harmful to living organisms, but it also has beneficial applications in medicine and industry.
Examples of Alpha Decay
Several heavy elements undergo alpha decay, including:
- Uranium-238: $ {}{92}^{238} U \longrightarrow {}{90}^{234} Th + {}_2^4 He $
- Radium-226: $ {}{88}^{226} Ra \longrightarrow {}{86}^{222} Rn + {}_2^4 He $
- Polonium-210: $ {}{84}^{210} Po \longrightarrow {}{82}^{206} Pb + {}_2^4 He $
These examples illustrate how alpha decay leads to the transformation of one element into another, with the emission of an alpha particle and the release of energy.
Conclusion
Alpha decay is a fundamental nuclear process driven by the instability of heavy nuclei. The emission of a stable alpha particle allows the nucleus to move towards a more stable configuration by reducing its mass and atomic number. Understanding alpha decay is crucial in various fields, including nuclear physics, nuclear chemistry, and nuclear medicine. The process has significant implications for the behavior of radioactive materials and the applications of nuclear technology.
In the intricate world of nuclear reactions, where atomic nuclei undergo transformations, the principles of conservation laws stand as unwavering pillars. These laws, fundamental to physics, dictate that certain physical quantities remain constant throughout a nuclear reaction. Among the most crucial are the conservation of mass number and the conservation of atomic number. These conservation laws not only govern the balancing of nuclear equations but also provide invaluable insights into the nature of nuclear processes and the identities of the participating nuclei. Understanding these laws is paramount for anyone venturing into the realms of nuclear chemistry and physics.
Conservation of Mass Number The Nucleon Count
The mass number, often denoted by the symbol 'A', represents the total number of nucleons within an atomic nucleus. Nucleons are the fundamental building blocks of the nucleus, encompassing both protons and neutrons. The conservation of mass number is a cornerstone principle in nuclear reactions, asserting that the total number of nucleons remains constant throughout the reaction. In other words, nucleons are neither created nor destroyed; they are merely rearranged.
This principle is elegantly expressed in the balancing of nuclear equations. The sum of the mass numbers on the reactant side (the initial nuclei) must precisely equal the sum of the mass numbers on the product side (the resulting nuclei and particles). Any discrepancy in the mass number balance would indicate an error in the equation or a misunderstanding of the reaction.
Illustrative Examples
Consider the alpha decay of Uranium-238 ($ {}_{92}^{238} U $):
$ {}{92}^{238} U \longrightarrow {}{90}^{234} Th + {}_2^4 He $
On the reactant side, the mass number is 238 (Uranium-238). On the product side, we have Thorium-234 (mass number 234) and an alpha particle ($ {}_2^4 He $, mass number 4). The sum of the mass numbers on the product side is 234 + 4 = 238, which perfectly matches the mass number on the reactant side, thus confirming the conservation of mass number.
Another example is the nuclear reaction where Nitrogen-14 is bombarded with an alpha particle:
$ {}_{7}^{14} N + {}2^4 He \longrightarrow {}{8}^{17} O + {}_1^1 H $
On the reactant side, the total mass number is 14 (Nitrogen-14) + 4 (alpha particle) = 18. On the product side, the mass number is 17 (Oxygen-17) + 1 (proton) = 18. Again, the mass number is conserved.
Implications and Applications
The conservation of mass number has profound implications. It allows us to predict the mass numbers of unknown nuclei produced in nuclear reactions. By carefully accounting for the nucleons on both sides of the equation, we can deduce the mass number of the missing product, providing a crucial clue to its identity.
This principle is also invaluable in nuclear transmutation, the process of converting one element into another. By understanding the conservation of mass number, scientists can design nuclear reactions to synthesize specific isotopes of elements, a cornerstone of nuclear medicine and materials science.
Conservation of Atomic Number The Proton Count
The atomic number, denoted by the symbol 'Z', is the defining characteristic of an element. It represents the number of protons within the nucleus. The conservation of atomic number is another inviolable principle in nuclear reactions, stating that the total number of protons remains constant throughout the reaction. This implies that the total positive charge within the nucleus is conserved.
Similar to the mass number, the atomic number must be balanced in nuclear equations. The sum of the atomic numbers on the reactant side must equal the sum of the atomic numbers on the product side. This conservation law is instrumental in identifying the elements involved in a nuclear reaction.
Illustrative Examples
Revisiting the alpha decay of Uranium-238:
$ {}{92}^{238} U \longrightarrow {}{90}^{234} Th + {}_2^4 He $
On the reactant side, the atomic number is 92 (Uranium). On the product side, we have Thorium (atomic number 90) and an alpha particle (atomic number 2, as it contains two protons). The sum of the atomic numbers on the product side is 90 + 2 = 92, confirming the conservation of atomic number.
Considering the nuclear reaction involving Nitrogen-14 and an alpha particle:
$ {}_{7}^{14} N + {}2^4 He \longrightarrow {}{8}^{17} O + {}_1^1 H $
On the reactant side, the total atomic number is 7 (Nitrogen) + 2 (alpha particle) = 9. On the product side, the atomic number is 8 (Oxygen) + 1 (proton) = 9. The atomic number is conserved.
Implications and Applications
The conservation of atomic number is essential for identifying the elements formed during nuclear reactions. If the atomic number of a product is unknown, we can deduce it by applying this conservation law. This allows us to determine the element's identity, as the atomic number uniquely defines an element.
Furthermore, this principle is vital in understanding the transformations that occur during radioactive decay. For instance, alpha decay always results in a decrease of two in the atomic number, indicating the formation of an element two places lower in the periodic table.
The Interplay of Conservation Laws
The conservation of mass number and the conservation of atomic number work synergistically to provide a comprehensive understanding of nuclear reactions. By simultaneously applying both principles, we can fully balance nuclear equations and identify all the participating nuclei and particles. These conservation laws are not merely mathematical tools; they reflect the fundamental nature of nuclear processes and the immutability of certain physical quantities.
Conclusion
The conservation of mass number and the conservation of atomic number are indispensable principles in the realm of nuclear chemistry and physics. They govern the balancing of nuclear equations, enable the identification of unknown nuclei, and provide insights into the mechanisms of nuclear reactions. These conservation laws are not just theoretical constructs; they are the bedrock upon which our understanding of the nucleus and its transformations rests. Mastering these principles is crucial for anyone seeking to unravel the mysteries of the atomic nucleus and harness the power of nuclear reactions.
In the provided nuclear equation, our mission is to identify the unknown element "X" produced during the alpha decay of Plutonium-239. To accomplish this, we embark on a step-by-step solution, meticulously applying the principles of conservation of mass number and atomic number. This journey will unveil the identity of element X and solidify our understanding of nuclear transformations.
Step 1 Write Down the Nuclear Equation
The nuclear equation serves as our roadmap, outlining the transformation that occurs. The given equation is:
$ {}_{94}^{239} Pu \longrightarrow X + {}_2^4 He $
This equation vividly depicts the alpha decay of Plutonium-239 ($ {}_{94}^{239} Pu {}_2^4 He $).
Step 2 Apply the Conservation of Mass Number
As we've learned, the conservation of mass number is a fundamental principle in nuclear reactions. It dictates that the total number of nucleons (protons and neutrons) remains constant throughout the reaction. In equation form:
Mass number of reactants = Mass number of products
Applying this to our equation:
239 (Pu) = A (X) + 4 (He)
Here, 239 is the mass number of Plutonium-239, A is the unknown mass number of element X, and 4 is the mass number of the alpha particle. Solving for A, we isolate the mass number of element X:
A = 239 - 4 = 235
Thus, we've determined that element X has a mass number of 235.
Step 3 Apply the Conservation of Atomic Number
The conservation of atomic number, another cornerstone principle, states that the total number of protons remains constant throughout a nuclear reaction. In equation form:
Atomic number of reactants = Atomic number of products
Applying this to our equation:
94 (Pu) = Z (X) + 2 (He)
Here, 94 is the atomic number of Plutonium, Z is the unknown atomic number of element X, and 2 is the atomic number of the alpha particle. Solving for Z, we determine the atomic number of element X:
Z = 94 - 2 = 92
Therefore, element X possesses an atomic number of 92.
Step 4 Identify Element X Using the Periodic Table
The periodic table, a meticulously organized chart of elements, is our indispensable tool for identifying element X. The periodic table arranges elements in ascending order of their atomic numbers. By locating the element with an atomic number of 92, we can unveil the identity of element X.
Consulting the periodic table, we find that the element with an atomic number of 92 is Uranium (U). This reveals that element X is Uranium.
Step 5 Specify the Isotope of Element X
We've identified element X as Uranium, but we must specify the isotope. Isotopes are variants of an element that share the same atomic number but differ in their mass numbers due to varying neutron counts. Element X has a mass number of 235, as determined in Step 2. Therefore, element X is Uranium-235 ($ {}_{92}^{235} U $).
Step 6 Write the Complete Nuclear Equation
Now that we've identified element X as Uranium-235, we can write the complete nuclear equation:
$ {}{94}^{239} Pu \longrightarrow {}{92}^{235} U + {}_2^4 He $
This equation comprehensively describes the alpha decay of Plutonium-239 into Uranium-235, accompanied by the emission of an alpha particle.
Step 7 Verify the Conservation Laws
As a final step, we rigorously verify that our solution adheres to the conservation laws. We must ensure that both the mass number and the atomic number are balanced on both sides of the equation.
- Mass Number Conservation: 239 (Pu) = 235 (U) + 4 (He) → 239 = 239 (Verified)
- Atomic Number Conservation: 94 (Pu) = 92 (U) + 2 (He) → 94 = 94 (Verified)
The mass number and atomic number are indeed conserved, validating our solution.
Conclusion
Through this methodical, step-by-step approach, we successfully identified element X as Uranium-235 ($ {}_{92}^{235} U $). We meticulously applied the conservation of mass number and atomic number, utilizing the periodic table as our guide. This exercise underscores the power of these conservation laws in deciphering nuclear reactions. By mastering these principles, we can confidently navigate the intricate world of nuclear transformations and unravel the identities of the nuclei involved.